The Biot-Savart Law

• Magnetic fields go around the wire – they areperpendicular to the direction of current

• Magnetic fields are perpendicular to the separation between the wire and the point where you measure it - Sounds like a cross product!

r

I

2

Id

rB

ds

ˆd s r

02

ˆ

4

I d

r

s rB 0

34

I d

r

s rB

70 4 10 T m/A

• Permeability of free space• The Amp is defined to work out this way

Ch. 30

=

Sample Problem 02

ˆ

4

I d

r

s rB

P

A loop of wire consists of two quarter circles of radii R and 2R, both centered at a point P, and connected with wires going

radially from one to the other. If a current I flows in the loop, what is the

magnetic field at the point P?

R 2R

• Do one side at a time• First do one of the straight segments• ds and r-hat are parallel

• No contribution to the integral• Other straight segment is the same

dsr̂ˆ 0d s r

dsr̂

• Now do inner quarter loop (assume x-y)

dsr̂

ˆˆd ds s r k0

inner 2ˆ

4

I ds

R

B k 02

2ˆ4 4

I R

R

k 0 ˆ8

I

R

k

• Outer loop opposite direction, similar

ds

r̂

0outer

ˆ16

I

R

B k

0 ˆ16

I

R

B k

I

Magnetic Field from a Finite Wire

r

I

ds

• Magnetic field from a finite straight wire:• Let a be the distance the point is from the wire• Let x be the horizontal separation

P

a

x

ˆ ˆx a r i j2 2r x a

-x1 x2

ˆd dxs i

2

1

03 22 2

ˆ ˆ ˆ

4

x

x

dx x aI

x a

i i jB

2

1

03 22 2

ˆ4

x

x

Ia dx

x a

k

0 2 1

2 2 2 22 1

ˆ

4

I x x

a x a x a

k

01 2cos cos

4

IB

a

21

11 2 2

1

cosx

x a

2

2 2 22

cosx

x a

Warning: My differs from that of the book

O

Magnetic Field from a Wire

I

• Magnitude is found from the formula• Direction is found from the right hand rule

• Place thumb in direction of current flow• Fingers curl in direction of B-field

a

01 2cos cos

4

IB

a

21

• Infinite wire:• Angles are simple

1 2

1 2

0

cos cos 1

0

2

IB

a

Two wires have the same current I flowing through them. If we want the magnetic field between them to be large and up, we should have the current in the upper one flow _____ and the lower one ______A) Left, Right B) Left, LeftD) Right, Left E) Right, Right

CT-1-A battery establishes a steady current around the circuit below. A compass needle is placed in the plane of the circuit successively just to the left of points P, Q, and R. The relative deflection of the needle, in de-scending order, is

A. P, Q, R. B. Q, R, P. C. R, Q, P. D. P, R, Q. E. Q, P, R

Ans E

Sample problemA regular hexagon whose center is a

distance a = 1 cm from the nearest side has current I = 4.00 A flowing around it. The

current flows N = 500 times around. What is the total magnetic field at the center?

I

a60

60

• Draw in the two directions from the center to the corners of one segment

• Top angle is one-sixth of a circle, or 60 degrees• Total angles in circle is 180, so other two angles are 60 each• Use formula to get magnetic field – right hand rule says up.

60

0 cos 60 cos 604

IB

a

0

4

I

a

• Multiply by all six side, and then by 500 cycles

0tot

6

4

INB

a

76 4 10 T m/A 4 A 500

4 0.01 m

0.120 T

Right Hand Rule for LoopsI

• If you curl your fingers in the direction the current flows, thumb points in direction of B-field inside the loop

• Works for solenoids too (later)

Curl fingers around I and Thumb point to Mag dipole moment.Also gives B inside a loop

B

Warmup 14

Force Between Parallel Wires

• One wire – infinite – creates a magnetic field• Other wire – finite or infinite – feels the force

I1

I2d

L0 1

2

IB

d

2 2I F L B0 1 2

2

I IF

L d

• Attractive if current is parallel, repulsive if anti-parallel

F

CT-2 Two parallel wires are connected to two separate DC power supplies so that the direction of their currents are initially parallel. When the leads to one of the wires are reversed

+-

+-

I1 I2

A. The wires move closer to each other B. The wires mover apart C. The wires do not move.

Ans B

CT-3- Consider two parallel wires carrying currents I1 and I2 respectively. The wires are a small distance a apart. Which of the following (is) are true:

A. If I1 = 2I2 and the directions of the currents are in the same direction, then the attractive force on the wire carrying I2 is 2 times that on the wire carrying I1.

B. If I1 = 2I2 and the directions of the currents are in the same direction, then the attractive force on the wire carrying I1 is 2 times that on the wire carrying I2.

C. If the magnitudes of the currents are the same but their directions are opposite to each other the magnetic field at a distance r > a is twice what it would be if only one wire were present.

D. If the magnitudes of the currents are the same but their directions are opposite to each other the magnetic field at a distance r > a is zero.

E. Two of the above F. None of the above [Don’t click]

Ans D

Warmup 14

Ampere’s Law (original recipe)• Suppose we have a wire coming out of the plane• Let’s integrate the magnetic field around a closed path

• There’s a new symbol for such an integral• Circle means “over a closed loop”

• The magnetic field is parallel to direction of integration

0

2

IB

a

dB s B ds 0 22

Ir

r

0I• What if we pick a different path?

dB s cosBds I0

2

Ird

r

0d I B s

ds

cosds rd

ds cosrd

• We have demonstrated this is true no matter what path you take• Wire don’t even need a straight infinite wire

• All that matters is that current passes through the closed Ampere loop

Understanding Ampere’s Law• If multiple currents flow through, add up all that are inside the loop• Use right-hand rule to determine if they count as + or –

• Curl fingers in direction of Ampere loop• If thumb points in direction of current, plus, otherwise minus

• The wire can be bent, the loop can be any shape, even non-planar

0d I B s

5 A

2 A1 A

4 A

7 A

There are currents going in and out of the screen as sketched at right. What is the ingtegral of the magnetic field around the path sketched in purple?A) 0(11 A) B) 0(-11 A) C) 0(3 A)D) 0(-3 A) E) None of the above

• Right hand rule causes thumb to point down• Downward currents count as +, upwards as –

0 4 A 2 A 5 Ad B s

JIT Quick Quiz 30.3Rank the magnitudes of ·d for the closed paths in the figure, from greatest to least.

Ans c>a>d>b

JIT Quick Quiz 30.4Rank the magnitudes of ·d for the closed paths in the figure, from greatest to least.

Ans a=c=d>b

Using Ampere’s Law• Ampere’s Law can be used – rarely – to calculate magnetic fields• Need lots of symmetry – usually cylindrical

A wire of radius a has total current I distribu-ted uniformly across its cross-sectional area.

Find the magnetic field everywhere.

I I

End-on view

• Draw an Ampere loop outside the wire – it contains all the current• Magnetic field is parallel to the direction of this loop, and constant

around it• Use Ampere’s Law• But we used a loop outside the wire, so we only have it for

r > a

0I d B s B ds 2 rB0

2

IB

r

Using Ampere’s Law (2)• Now do it inside the wire• Ampere loop inside the wire does not contain all the current• The fraction is proportional to the area

End-on view

2 rB

2

2rI r

I a

a2 2rI Ir a

0 rI d B s0

2rI

Br

022

Ir

a

02

0

2

2

I r r aB

Ir a r a

Warmup 15

Solenoids• Consider a planar loop of wire – any

shape – with a current I going around it• Now, stack many, many such loops

• Treat spacing as very closely spaced• Assume stack is tall compared to size

of loop• Can show using symmetry that magnetic

field is only in vertical direction• Can use Ampere’s Law to show that it is

constant inside or outside the solenoid

00 I d B s 1B L 0 2B L 0

1 2B B

• But magnetic field at infinity must be zero

outside 0B

Field Inside a Solenoid• It remains only to calculate the magnetic field inside• We use Ampere’s law

• Recall, no significant B-field outside• Only the inside segment contributes

• There may be many (N) current loops within this Ampere loop

• Let n = N/L be loops per unit length

0 totI d B s

L

inB LtotI NI

0in

NIB

L

in 0B nI

• Works for any shape solenoid, not just cylindrical• For finite length solenoids, there are “end

effects”• Real solenoids have each loop connected to the

next, like a helix, so it’s just one long wire

Magnetic Flux• Magnetic flux is defined exactly the same way

for magnetism as it was for electricityˆ

B dA B n

A cylindrical solenoid of radius 10 cm has length 50 cm and has 1000 turns of wire going around it. What is the magnetic field inside it, and the magnetic flux through it, when a

current of 2.00 A is passing through the wire?

in 0B nI 7 1000 2 A4 10 T m/A

0.5 m

in 0.00503 TB

B BA 2B R 20.00503 T 0.1 m

4 21.579 10 T mB A Tesla meter2 is also

called a Weber (Wb)

Gauss’s Law for Magnetism• Magnetic field lines always go in circles – there are no magnetic

monopole sources• For closed surfaces, any flux in must go out somewhere else

A regular tetrahedron (four sides, all congruent) has a cylin-drical magnet placed in the middle of the bottom face. There is a total of 0.012 Tm2 of magnetic flux entering the bottom face. What is the total flux from one of the three top faces?A) 0.006 T m2 B) 0.004 T m2 C) 0.003 T m2

D) 0.012 T m2 E) None of the above

ˆ 0B dA B n

• Flux in bottom must equal total flux out other three sides• Other three sides must have equal flux, by symmetry

21 2 3 0.012 T m 2

13 0.012 T m

CT-4 - A sphere of radius R is placed near a long, straight wire that carries a steady current I. The magnetic field generated by the current is B. The total magnetic flux passing through the sphere is

A. o I. B. o I/(4 R 2 ). C. 4 R 2/ o I. D. zero. E. need more information

Ans D

Ex – Serway 30.45 A cube of edge length = 2.50 cm is positioned as shown below. A uniform magnetic field given by B = (5.00 i + 4.00 j + 3.00 k) T exists throughout the region. (a) Calculate the flux through the shaded region. (b) What is the total flux through the six faces?

Solve on Board

Warmup 15

http://www.youtube.com/watch?v=A1vyB-O5i6E